Offset logarithmic integral inequality

Prove that

$\int_2^{e+1} \frac{\mathrm{d}t}{\ln t} < e$

Solution

We have successively:

$\begin{align*} \hspace{-1em}\ln x \leq x-1 &\Rightarrow \ln \frac{1}{x} \leq \frac{1}{x}-1 \\ &\Rightarrow -\ln x \leq \frac{1}{x}-1 \\ &\Rightarrow \ln x \geq 1-\frac{1}{x} \\ &\Rightarrow \frac{1}{\ln x} \leq \frac{1}{1-\frac{1}{x}} \\ &\Rightarrow \int_{2}^{e+1} \frac{\mathrm{d}t}{\ln t} < \int_2^{e+1} \frac{\mathrm{d}t}{1-\frac{1}{t}} \\ &\Rightarrow \mathbf{\int_{2}^{e+1} \frac{\mathrm{d}t}{\ln t} < e} \end{align*}$

Isomorphic groups

Inverse zeta(3) limit

Pell - Lucas series

Inequality of a function

A contour integral

The group is abelian

An interesting limit

Uniform convergence

Explicit formula of f

One thought on “Offset logarithmic integral inequality”

Can you improve the inequality by actually showing that

$\int_2^{e+1} \frac{\mathrm{d}t}{\ln t} < e-1$

I do not have an answer to this question.